3 Answers
3

Since the real and imaginary parts are very much dependent on one another (if you have the value of one, you know the value of the other exactly), it seems like you could apply the marginal pdf of the real part $r$, given a value of the imaginary part $i$:

$$
f_{ri}(r, i) = f_{r | i}(r\ |\ i) f_i(i)
$$

You noted the pdf of the real and imaginary parts individually:

$$
f_r(z) = f_i(z) = \frac{1}{\pi\sqrt{1-z^2}}
$$

That leaves the marginal pdf $f_{r | i}(r\ |\ i)$. Remember that for a given realization of the random variable $Z$, the two components are deterministically related:

$$
r^2 + i^2 = \cos^2(\theta) + \sin^2(\theta) = 1
$$

Given this relationship, we can solve for $r$ in terms of $i$:

$$
r^2 = 1 - i^2
$$$$
r = \pm \sqrt{1-i^2}
$$

Therefore, the marginal pdf of $r$ given a value of $i$ is a pair of impulses:

Thinking about this geometrically, for every horizontal line $i = i_0$ (for $i_0 \in [-1, 1]$) in the $ri$ plane, there are only two points $r_0 = \pm \sqrt{1 - i_0^2}$ that are nonzero, and the pdf has has infinite height at those points. As we might expect, those points of intersection (i.e. points where the pdf is nonzero) are where the horizontal line intersects with the unit circle!

This means that the joint pdf is zero-valued, except along the unit circle, where it takes on infinite height. That aligns with intuition, as the definition of the random variable $Z$ ensures that can only take on values that are on the unit circle.

There's nothing special about the specific way I laid this out; you could also transpose the problem and look at vertical lines in the $ri$ plane of the form $r = r_0$ and you would find the same relationship due to the close coupling of the two random variables.

I believe this formulation is equivalent to that in AlexTP's answer, but his derivation is probably more intuitive.

$\begingroup$is there a typo in $r = 1- i^2$ ; possibly you mean $r^2=1-i^2$ ? Also can we show that the joint pdf integrates to 1 in x-y (or eqv r-i) plane. Again also the joint pdf $f_{ri}(r,i)$ is not circularly symmetric? (the denominator?). Cannot we simply use the logic that given $z = e^{j \theta} = \cos(\theta) + j \sin(\theta) = x + jy$ with $\theta$ uniform in [$0,2\pi$] and the constraint $x^2+y^2=1$ simply offers a circularly symmetric joint pdf of the type $f_{xy}(x,y) = K \delta_2(x^2+y^2-1)$ or $f_{R,\theta} = K \delta_2(R-1)$ ; a ring impulse on x-y plane? where K is possibly $1/2\pi$ ?$\endgroup$
– Fat32Nov 14 '18 at 22:57

$\begingroup$@Fat32: I made some fixes that led me to what sounds like what you were getting at in your comment above. AlexTP's answer still is probably more intuitively pleasing though.$\endgroup$
– Jason RNov 15 '18 at 1:50

1

$\begingroup$There is a typo in copying $f_{r|i}(r|i)$ to $f_{r,i}(r,i)$. And I observe that $f_{r,i}(0,1) = \infty$ is different from $f_{r,i}(1,0) = \frac{1}{2\pi}$. I think they should have the same value, shouldn't they?$\endgroup$
– AlexTPNov 16 '18 at 0:14

Avoid complicated calculations, let $X$ and $Y$ be i.i.d. standard normal random variables, your random variable $Z$ has the same distribution of $V$$$V \triangleq \left(\frac{X}{\sqrt{X^2+Y^2}},\frac{Y}{\sqrt{X^2+Y^2}} \right)$$
(easy to see $\lVert V\rVert=1$ and the angle of $V$ is equivalent to the angle of a circularly symmetric Normal hence uniform).

$\begingroup$The intuition seems to be so simple behind the proof. All you need to do is pick points from the surface of a unit sphere :)$\endgroup$
– MaxtronNov 15 '18 at 0:14

$\begingroup$Can't we just remove the $||z||$ in the denominator? Otherwise, how would you evaluate the expression for $||z||\rightarrow 0$? Of course, we would like $f_Z(0)=0$ to hold.$\endgroup$
– Matt L.Nov 15 '18 at 9:57

$\begingroup$@MattL. you are right, it is wrong having $\lVert z \rVert$ in the denominator, not due to the evaluation at limits but because it is simply wrong. I have fixed the final result.$\endgroup$
– AlexTPNov 15 '18 at 10:29

1

$\begingroup$@AlexTP: I still think that there shouldn't be any variable in the denominator. Have a look at my answer and please tell me if/where I go wrong.$\endgroup$
– Matt L.Nov 15 '18 at 16:31

Based on the existing answers, which opened my eyes for what's going on here, I would like to present yet another very simple expression for the solution, which is just slightly different from the one in AlexTP's answer (and which turned out to be equivalent to the one given in Jason R's answer, as shown below in the EDIT-part).

[EDIT: now that AlexTP has edited his answer, our expressions for the PDF are identical; so all three answers finally agree with each other].

Let the complex random variable $Z=X+jY$ be defined as

$$Z=\rho e^{j\theta}\tag{1}$$

where the radius $\rho$ is deterministic and given, whereas the angle $\theta$ is random and uniformly distributed on $[0,2\pi)$. I state without further proof that $Z$ is circularly symmetrical, from which it follows that its probability density function (PDF) must satisfy

Since the PDF must be zero everywhere except for $r=\rho$, and since it must integrate to unity (when integrated over the 2-dimensional plane), the only possible PDF is

$$f_Z(r)=\frac{1}{2\pi}\delta(r-\rho)\tag{3}$$

It can be shown that $(3)$ leads to the correct marginal densities for the random variables $X$ and $Y$.

EDIT:

After some very useful discussion in the comments it appears that we've managed to agree on one solution to the problem. I will show in the following that the unassuming formula $(3)$ is actually equivalent to the more involved looking formula in Jason R's answer. Note that I use $r$ for the magnitude (radius) of the complex RV $Z$, whereas in Jason's answer $r$ denotes the real part of $Z$. I will use $x$ and $y$ for the real and imaginary parts, respectively. Here we go:

$\begingroup$@AlexTP: Take as an extreme example $\rho=0$, which makes the RV $Z$ deterministic, it's always zero. It's PDF should then be a Dirac impulse at $r=z=y=0$, which is what I get with the formula I suggested. With the formula in your answer we get an undefined term in that case.$\endgroup$
– Matt L.Nov 15 '18 at 16:58

$\begingroup$it's true, but I think the value of PDF can be infinity and only the CDF should be bounded. In other words, the Dirac impulse matters only inside integrals.$\endgroup$
– AlexTPNov 15 '18 at 17:43

$\begingroup$Yes, but if you integrate your pdf (for $\rho=0$), what do you get (and how would you do it)?$\endgroup$
– Matt L.Nov 15 '18 at 17:48

$\begingroup$I would do like this $$\lim_{\rho \to 0^+}\int_\theta \int_{r=0}^{+\infty}\frac{1}{2\pi\rho}\delta(r-\rho) r dr d\theta = \lim_{\rho \to 0^+} \int_\theta \frac{1}{2\pi\rho} \rho d\theta = \int_\theta \frac{1}{2\pi} d\theta = 1$$ The integral is finite while the PDF is not thanks to the Jacobian determinant. My point is that you need that Jacobian determinant $r$ whenever you integrate the PDF in polar coordinate. Tell me if you find it unconvinced.$\endgroup$
– AlexTPNov 15 '18 at 20:50